Net force acting on particle of mass 2 kg is 10N in north direction.At t=o, particle was moving eastwards with 10m/s. Find displacement and velocity of particle after 2 seconds.

Ans)
given force =F=10N

mass M=2kg

velocity u=10m/s

then acceleration a=F/M=10/2=5m/s^2

then displacement=s=ut+1/2at^2

here u=initial velocity

t=time

a=acceleration

then we get the displacement is=30m

2.velocity=u+at

then we get velocity as 20m/s.

Posted Date: 3/8/2013 7:08:50 AM | Location : United States

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Initial velocity is 10 m/s in x-direction. Now a force of 10N acts in y-direction  So , using F = ma                 a = f/m = 10/2 = 5m/s2in y-direction  After 2 seconds velocity in y-direction would become 10 m/s Now we can take the resultant of the two velocities i.e 10 m/s in x-direction & 10m/s in y-direction Therefore Resultant velocity = (102+102)1/2                     = (200)1/2 and 450 North of East Displacement = velocity x time                      = (200)1/2x 2                     = 20x(2)1/2m 450 north of east Posted by | Posted Date: 3/8/2013 7:09:55 AM

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